This course explains how to analyze circuits that have direct current (DC) current or voltage sources. A DC source is one that is constant. Circuits with resistors, capacitors, and inductors are covered, both analytically and experimentally. Some practical applications in sensors are demonstrated.

Taught By

Dr. Bonnie H. Ferri

Professor

Dr. Joyelle Harris

Director of the Engineering for Social Innovation Center

Dr Mary Ann Weitnauer

Transcript

[MUSIC] Welcome back to linear circuits, this is Doctor Weitnauer. This lesson is finding initial conditions in second order circuits. Our objective are to find out what type of initial conditions are needed, and apply circuit analysis to calculate them. This builds on many things. A second order circuit is modeled by a second order differential equation. The I V relationships for capacitors and inductors. The current inductor is continuous in time. The voltage across a capacitor is continuous in time. And finally, KCL. Our motivation is, we need these initial conditions in the last step of finding the complete response of a circuit with reactive elements. What initial conditions are needed? If we want to find y(t) for t greater than or equal to 0, when y(t) satisfies a differential equation such as this. Because this is a second order differential equation, we need y(0+) and its derivative at 0+. How do you normally start? We are usually given the initial voltage across the capacitor or the initial current through the inductor. Or if we're not given those things, then we can derive them by DC analysis prior to t=0, and use their continuity properties. But even if you have these, we are not done, we have to find that derivative and this usually requires circuit analysis. Here's a simple example, the series RLC circuit. Find the initial conditions for the solution for v(t). Suppose we were given v(0+) and i, i being the current through the inductor at times 0+. However, because we are trying to find the initial conditions for the solution for v(t), this is not all we need. We have one part of it, but we need the derivative of v(0+). So we can use i = C dv dt, and then solve for dv dt at 0+. And that will give us 1 over C times i(0+) which we are given. This is a harder example. We will determine the initial conditions needed for v(t), t greater than or equal to 0, assuming the circuit is at rest, that is, DC steady state, prior to t=0. To do this, we should consider what this circuit looks like prior to t=0. There are four things that we need to consider. This resistor will not play a role because the switch is open. The capacitor will behave as an open circuit, the inductor will behave as a short circuit. And the voltage, because u(t) will be 0 for negative time, the voltage is only 7 volts. So this is what our simplified circuit looks like for t<0. First we can notice that the resistor is in parallel with the voltage source. So that means that v(0-) is 7 volts because of the continuity of voltage in capacitors, that is also v(0+). There's one of the initial conditions that we need. The current in the inductor, we can find that by using Ohm's law. And because of the continuity of current as a function of time in an inductor, that is also the current at i(0+). But that is not the second initial condition we need, we need this derivative at t=0+. Now, I want to emphasize this is at 0+. The circuit is different for positive time, so we need to redraw the circuit for t>0. Here is that circuit, and we see that we can do a little more simplification by combining these parallel resistors into one effective resistance of R over 2. Now I want to make a little room on the screen and we'll continue. Note that we have these initial values of the voltage v and the current i through the inductor, and our goal is to find the derivative of the voltage at t = 0+. It's going to be helpful for us to define two currents. i sub r through that branch, and i sub c through this branch, and we'll be using KCL, which is, i = ir + ic. This is that circuit analysis I was telling you about. Now we'll going to use the ic = c dv dt. That is why I've chosen to use KCL. It allows me to bring in that derivative that we're seeking. I'm also going to note that ir, because it's in parallel with the capacitor, we can find it by saying it's V over the value of the resistance, which is R over 2. Next I will solve the KCL equation for ic, and then divide by c, giving me dv dt = 1 over c [i- ir]. And this is going to be true for anytime, but in particular we want it for t = 0+. That's the other initial condition we need, and we will be able to just plug in the current through the inductor at i+ and then we can use the V(0+) divided by (R/2). And that gives us the other initial condition that we need. Okay, to summarize the key concepts in this lesson. Finding the initial conditions for a second order circuit requires more than just finding the inductor currents and the capacitor voltages. In general, you're going to have to do some circuit analysis to find the derivative of the desired variable at t = 0+. And it's important to redraw the circuit for t>0 when you're trying to find that derivative. Thank you. [MUSIC]

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